Double 的 BigDecimal 值不正确？

发布于 2024-09-18 09:01:36 字数 428 浏览 6 评论 0原文

我正在尝试从字符串创建 BigDecimal。别问我为什么，我就是需要！这是我的代码：

Double theDouble = new Double(".3");
System.out.println("The Double: " + theDouble.toString());
BigDecimal theBigDecimal = new BigDecimal(theDouble);
System.out.println("The Big: " + theBigDecimal.toString());

这是我得到的输出？

The Double: 0.3
The Big: 0.299999999999999988897769753748434595763683319091796875

有什么想法吗？

原文

I am trying to make a BigDecimal from a string. Don't ask me why, I just need it! This is my code:

Double theDouble = new Double(".3");
System.out.println("The Double: " + theDouble.toString());
BigDecimal theBigDecimal = new BigDecimal(theDouble);
System.out.println("The Big: " + theBigDecimal.toString());

This is the output I get?

The Double: 0.3
The Big: 0.299999999999999988897769753748434595763683319091796875

Any ideas?

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浪漫之都 2024-09-25 09:01:36

当您创建双精度型时，无法准确表示值 0.3。您可以从不带中间 double 的字符串创建 BigDecimal，如

new BigDecimal("0.3")

浮点数表示为二进制分数和指数。因此，有些数字无法准确表示。在以 10 为基数的数字中也存在类似的问题，例如 1/3，即 0.333333333..... 1/3 的任何十进制表示都是不精确的。这种情况发生在二进制分数的不同集合中，0.3 是二进制中不精确的分数集合之一。

When you create a double, the value 0.3 cannot be represented exactly. You can create a BigDecimal from a string without the intermediate double, as in

new BigDecimal("0.3")

A floating point number is represented as a binary fraction and an exponent. Therefore there are some number that cannot be represented exactly. There is an analogous problem in base 10 with numbers like 1/3, which is 0.333333333..... Any decimal representation of 1/3 is inexact. This happens to a DIFFERENT set of fractions in binary, and 0.3 is one of the set that is inexact in binary.

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长安忆 2024-09-25 09:01:36

另一种方法是使用 MathContext.DECIMAL32 保证 7 位精度（在我们的例子中已经足够好了）：

Double theDouble = new Double(".3");
System.out.println("The Double: " + theDouble.toString());
BigDecimal theBigDecimal = new BigDecimal(theDouble, MathContext.DECIMAL32);  // <-- here
System.out.println("The Big: " + theBigDecimal.toString());

OUTPUT

The Double: 0.3
The Big: 0.3000000

Another way is to use MathContext.DECIMAL32 which guarantees 7 digit precision (which is good enough in our case):

Double theDouble = new Double(".3");
System.out.println("The Double: " + theDouble.toString());
BigDecimal theBigDecimal = new BigDecimal(theDouble, MathContext.DECIMAL32);  // <-- here
System.out.println("The Big: " + theBigDecimal.toString());

OUTPUT

The Double: 0.3
The Big: 0.3000000

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惯饮孤独 2024-09-25 09:01:36

由于 new Double(".3") 无法准确表示，因此最接近的值为 0x1.3333333333333P-2 或 .299999999999999988897769753748434595763683319091796875，需要的是：

Double theDouble = new Double(".3");
System.out.println("The Double: " + theDouble.toString());
BigDecimal theBigDecimal = new 
BigDecimal(theDouble).setScale(2, RoundingMode.CEILING);  // <-- here
System.out.println("The Big: " + theBigDecimal.toString());

这将打印：

The Double: 0.3
The Big: 0.30

Since new Double(".3") can't be represented exactly, the nearest value is 0x1.3333333333333P-2 or .299999999999999988897769753748434595763683319091796875, what would be need to is this:

Double theDouble = new Double(".3");
System.out.println("The Double: " + theDouble.toString());
BigDecimal theBigDecimal = new 
BigDecimal(theDouble).setScale(2, RoundingMode.CEILING);  // <-- here
System.out.println("The Big: " + theBigDecimal.toString());

This will print:

The Double: 0.3
The Big: 0.30

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星軌x 2024-09-25 09:01:36

您可以为大数小数指定指定的精度。例如附加到您的示例：

Double theDouble = new Double(".3");
theBigDecimal = new BigDecimal(theDouble, new MathContext(2));
System.out.println("The Big: " + theBigDecimal.toString());

这将打印出“0.30”

You can give a big decimal a specified precision. e.g. append to your example:

Double theDouble = new Double(".3");
theBigDecimal = new BigDecimal(theDouble, new MathContext(2));
System.out.println("The Big: " + theBigDecimal.toString());

This will print out "0.30"

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